# proof of limit of nth root of n

In this entry, we present a self-contained, elementary
proof of the fact that ${lim}_{n\to \mathrm{\infty}}{n}^{1/n}=1$.
We begin by with inductive proofs of two integer
inequalities^{} — real numbers will not enter until
the very end.

###### Lemma 1.

For all integers $n$ greater than or equal to $\mathrm{5}$,

$${2}^{n}>{n}^{2}$$ |

###### Proof.

We begin with a few easy observations. First, a bit of arithmetic:

$${2}^{5}=32>25={5}^{2}$$ |

Second, some algebraic manipulation of the inequality $n>4$:

$n-1$ | $>3$ | ||

${(n-1)}^{2}$ | $>9$ | ||

${(n-1)}^{2}$ | $>2$ | ||

${n}^{2}-2n+1$ | $>2$ | ||

$2{n}^{2}$ | $>{n}^{2}+2n+1$ | ||

$2{n}^{2}$ | $>{(n+1)}^{2}$ |

These observations provide us with the makings of an inductive proof. Suppose that ${2}^{n}>{n}^{2}$ for some integer $n\ge 5$. Using the inequality we just showed,

$${2}^{n+1}=2\cdot {2}^{n}>2{n}^{2}>{(n+1)}^{2}.$$ |

Snce ${2}^{5}>{5}^{2}$ and ${2}^{n}>{n}^{2}$ implies that ${2}^{n+1}>{(n+1)}^{2}$ when $n\ge 5$ we conclude that ${2}^{n}>{n}^{2}$ dor all $n\ge 5$. ∎

###### Lemma 2.

For all integers $n$ greater than or equal to $\mathrm{3}$,

$${n}^{n+1}>{(n+1)}^{n}$$ |

###### Proof.

We begin by noting that

$${3}^{4}=81>64={4}^{3}.$$ |

Next, we make assume that

$${(n-1)}^{n}>{n}^{(n-1)}.$$ |

for some $n$. Multiplying both sides by $n$:

$$n{(n-1)}^{n}>{n}^{n}.$$ |

Multiplying both sides by ${(n+1)}^{n}$ and making use of the identity $(n+1)(n-1)={n}^{2}-1$,

$$n{({n}^{2}-1)}^{n}>{n}^{n}{(n+1)}^{n}.$$ |

Since ${n}^{2}>{n}^{2}-1$, the left-hand side is less than ${n}^{2n+1}$, hence

$${n}^{2n+1}>{n}^{n}{(n+1)}^{n}.$$ |

Canceling ${n}^{n}$ from both sides,

$${n}^{(n+1)}>{(n+1)}^{n}.$$ |

Hence, by induction, ${n}^{(n+1)}>{(n+1)}^{n}$ for all $n\ge 3$. ∎

###### Theorem 1.

$$\underset{n\to \mathrm{\infty}}{lim}{n}^{1/n}=1$$ |

###### Proof.

Consider the subsequence where $n$ is a power of $2$. We then have

$${({2}^{m})}^{(1/{2}^{m})}={2}^{m/{2}^{m}}.$$ |

By lemma 1, $$ when $m\ge 5$. Hence, $$. Since ${lim}_{m\to 0}{2}^{1/m}=1$, and ${({2}^{m})}^{1/{2}^{m})}>1$, we conclude by the squeeze rule that

$$\underset{m\to 0}{lim}{({2}^{m})}^{1/{2}^{m}}=1.$$ |

By lemma 2, the sequence $\{{n}^{1/n}\}$ is decreasing. It is clearly bounded from below by $1$. Above, we exhibited a subsequence which tends towards $1$. Thus it follows that

$$\underset{n\to \mathrm{\infty}}{lim}{n}^{1/n}=1.$$ |

∎

Title | proof of limit of nth root of n |
---|---|

Canonical name | ProofOfLimitOfNthRootOfN |

Date of creation | 2014-02-28 7:21:31 |

Last modified on | 2014-02-28 7:21:31 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 19 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 30-00 |

Classification | msc 12D99 |